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The network shown in FIG. A time at t = 0 starts transmitting the file to a 2Mbits C; B at t = 0.1 + e s (e infinitely small positive real number close to 0) is sent to the file 1Mbits D. Ignoring propagation delay and processing delay node.
please answer the following question:
1) If the figures using the network storage - packet switching and forwarding method, then A file will be delivered to the C 2Mbits how long? B 1Mbits files will be delivered to the D How long?
2) If the network uses stored figures - and-forward packet-switched, packet length of equal length 1 kbits, and ignores the packet header overhead and overhead packet disassembly, then A will be delivered to 2Mbits files take approximately how long C ? B 1Mbits files will be delivered to D takes approximately how long?
3) Packet switching compared with packet switching, which exchange more equitable? (I.e. small with less amount of data transmission, data transmission with long amount)
solution:
Packet switching: source sends the whole message.
Packet switching: When split into a series of relatively small data packets, a plurality of sources to share a router (in / out) link, shared link demand (statistical multiplexing).
Furthermore: In this problem, the storage - and-forward network router can simultaneously receive and transmit data for multiple users. As long as the interfaces are different, it can be received simultaneously. Similarly, packets may be transmitted simultaneously in different interfaces.
(1) message exchange
- Store and forward node is located to the left of E, the node on the right to F.
- AE overall packet transmission time required (to 2 Mbits / 10Mbits / S) S = 0.2 ; transfer is completed in 0.2 S .
- BE packet transmission time required for the overall (1Mbits / 10Mbits / S) = 0.1s ; complete transfer to (0.2 + E) S .
- A first packet transmission EF, the time required (to 2 Mbits / 20Mbits / S) = 0.1s ; been transmitted to 0.3s .
- EF B starts to transmit the packet at the end of the packet A, the time required (1Mbits / 20Mbits / S) = 0.05s ; complete transfer to (0.35 + E) S .
- FC transmission time required (to 2 Mbits / 10Mbits / S) S = 0.2 , the transfer is complete at 0.5s .
- FD required transmission time (1Mbits / 10Mbits / S) S = 0.1 , the transfer is complete to (0.45 + E) S .
In summary, A to C delivery 2Mbits need 0.5s . B to D delivery 1Mbits need 0.35s .
(2) The packet length for transmission of equal length 1kbits.
EF bandwidth is 20Mb, it will not be congestion problems.
Spending time should be divided by the size of the bottleneck link bandwidth whole message, together with the last departure from the router to the second time package.
In the process of transmitting BD, the AC also still being transmitted, the course of the last packet of the BD, it only enjoy 10M bandwidth, 0.1002s in the middle of two routers.
And the last packet of the AC energy of the intermediate exclusive 20M bandwidth, 0.20015s.
therefore
AC requires 0.2 + 0.001 + 0.0005 = 0.20015s
BD needs 0.1 + 0.001 + 0.0005 = 0.1002s
(3) 1, 2 can be seen that in both cases, less time packet switching. And in the case of packet switching, when a BD 0.1 + e s slow start transmitting on the link EF wait A packet transmission is completed additional 0.1s. In the case of packet switching, can be shared as needed to multiplex link, such message packet B is not blocked.
Therefore, packet switching more equitable.
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